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Find the sum of all the solutions of the equation $|2x^2+x-1|=|x^2+4x+1|$

Though I tried to solve it in desmos.com and getting the requisite answer but while solving it manually it is getting very lengthy.

I tried to construct the two parabola and mirror image the region below y axis but still getting it is getting complicated.

Is there any easy method to solve it and get the sum of all the solutions ?

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closed as off-topic by uniquesolution, Ak19, Nosrati, Peter Foreman, Adrian Keister Jul 2 at 17:58

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    $\begingroup$ Vieta's formula. $\endgroup$ – Eclipse Sun Jul 1 at 13:02
  • $\begingroup$ Note: $0$ is a root $\endgroup$ – J. W. Tanner Jul 1 at 13:03
  • $\begingroup$ @EclipseSun We can use Vieta's formula, but to solve a quadratic equation should also not be a real challenge. $\endgroup$ – Peter Jul 1 at 13:27
  • $\begingroup$ How can a question showing an effort AND context receive $4$ close-votes and $3$ down-votes ? $\endgroup$ – Peter Jul 1 at 13:30
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We are asked for the sum of the roots; we don't necessarily have to find the roots.

Squaring, we get $(2x^2+x-1)^2=(x^2+4x-1)^2$.

So $4x^4+4x^3...=x^4+8x^3...$.

So $\color{blue}3x^4-\color{blue}4x^3....=0$.

By Vieta's formulas, the answer is $\dfrac43$.

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  • $\begingroup$ inspired by Dr. Sonnhard Graubner (squaring) and Eclipse Sun (Vieta's formulas) $\endgroup$ – J. W. Tanner Jul 1 at 13:22
  • $\begingroup$ You can even do with $4x^4+4x^3+\cdots=x^4+8x^3+\cdots$. $\endgroup$ – Yves Daoust Jul 1 at 13:52
  • $\begingroup$ Yes @YvesDaoust; I thought of that after I posted $\endgroup$ – J. W. Tanner Jul 1 at 13:54
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Hint : Solve the two equations $$2x^2+x-1=x^2+4x+1$$ and $$2x^2+x-1=-x^2-4x-1$$

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  • $\begingroup$ And the conditions for that? $\endgroup$ – Dr. Sonnhard Graubner Jul 1 at 13:01
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    $\begingroup$ Since we have $|a|=|b|$ if and only if $a=b$ or $a=-b$ exactly the solutions of those equations solve the given equation. $\endgroup$ – Peter Jul 1 at 13:04
  • $\begingroup$ Better is, i think, to square both sides of the equation. $\endgroup$ – Dr. Sonnhard Graubner Jul 1 at 13:07
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    $\begingroup$ @Dr.SonnhardGraubner This gives an equation of degree $4$. I prefer this approach. $\endgroup$ – Peter Jul 1 at 13:15
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    $\begingroup$ @YvesDaoust I know because only the sum has to be determined. But my approach is already easy enough and answers the question whether there is an easy solution. $\endgroup$ – Peter Jul 1 at 13:54
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The expressions between the absolute value bars have the same or opposite signs. Hence there are two independent cases (by addition and subtraction):

$$3x^2+5x=0$$ and $$x^2-3x-2=0.$$

Then by Vieta,

$$-\frac53+3.$$


For complete rigor, one should show that no root is repeated. This is true, because the polynomials have no double root, and their $\text{gcd}$ is $1$.

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$$ |2x^2+x-1|=|x^2+4x+1|\\ (2x^2+x-1)^2=(x^2+4x+1)^2\\ (2x^2+x-1)^2-(x^2+4x+1)^2=0\\ [(2x^2+x-1)+(x^2+4x+1)]\cdot[(2x^2+x-1)-(x^2+4x+1)]=0\\ (3x^2+5x)\cdot(x^2-3x-2)=0\\ x\cdot(3x+5)\cdot(x^2-3x-2)=0\\ Solving\space for\space all\space cases,\space we\space get:\\ x=0\\ x=-\frac{5}{3}\\ x=\frac{3\pm\sqrt{17}}{2} $$

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Squaring and factorizing we get $$x(3x+5)(x^2-3x-2)=0$$ The solutions are given by $$x=-\frac{5}{3}\lor x=0\lor x=\frac{1}{2} \left(3-\sqrt{17}\right)\lor x=\frac{1}{2} \left(3+\sqrt{17}\right)$$

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    $\begingroup$ And the conditions for that ? $\endgroup$ – Yves Daoust Jul 1 at 13:06

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